Archive

Archive for the ‘Starters’ Category

Envelope puzzle and neat little square

June 25, 2020 Leave a comment

I’ve been looking through my saved puzzles again and I found this nice little one in the maths newsletter from Chris Smith (@aap03102):

It’s a nice little question that took me some thinking about.

First I considered the half squares with hypotenuse 2. As these are isoceles RATs, that means their side length is rt2 so each has an area of 1.

Then I thought about the half square with hypotenuse 3. Again it’s an isoceles RAT so pythagoras’s theorem gives us a sidelength of 3/rt2. So an area of 9/4.

These 3 half squares add to 17/4. The area of the rectangle is 6 so the part not covered by the half squares must be 7/4.

When thinking about the overlap, we need to consider what that means. I assume it means the bit that goes over the others, so in this case 9/4 (area of the halfsquare) – 7/4 (area of the gal). Which is 1/2 cm².

I think this is an amazing little question that I cant wait to try out in class. It also got me thinking about the square with diagonal of 2 and area of 2. It’s an interesting square really, and the only one where we see this. Consider a square, side length, x. The diagonal is (2x²)^½, the area is x². If we equate the. And square both sides we have 2x² = x⁴ , or x⁴ – 2x² = 0, so x²(x² – 2) = 0. This generates a solution when x = 0 (which is trivial and discountable as no square can exist without side length. We also get positive and negative square roots if 2. But we can ignore the negative as lengths in this case are scalar, so we have one answer. It’s a neat little square.

1-9 math walk

June 24, 2020 Leave a comment

Today I want to look at another puzzle I found on math walks (from Traci Jackson @traciteacher):

I love these 1-9 puzzles, and thought I’d have have a crack.

First I considered the 9, with the 1 gone already that means that the 9 must share a line with the 2 and the 3 to make 14.

That means that the 4 shares 1 line with a 2 and one with a 3. That means the 4 is one lines 4,2,8 and 4,3,7.

I considered these lines:

If I put the 7 on the left, I’d need the 6 at the intersection of the green lines. That would also mean that the 2 was above it, but I’d need another 6 on that line which doesn’t work.

So the 8 must be on the left and if we follow through we get:

I enjoyed this puzzle, if you have any cool 1-9 puzzles do send them on.

Categories: #MTBoS, KS2, KS3, Maths, Starters Tags: , , ,

A surprising fraction?

June 23, 2020 Leave a comment

The other day I saw this tweet from John Rowe (@MrJohnRowe):

I looked at the picture and decided the answer was probably a half. And thought about it a bit more. The 4 small circles in the top left should be the same as the white quarter circle below, and the 4 quarter circles in the top right should fit over the white circle below them. This was interesting to me, and I thought I’d look at the algebra.

Looking at the top left corner, the circles involved there have the smallest radius, so we will call that radius r. That means each circle has a radius of pi r².

Below it we have a square the same size and a quarter circle radius 4r, so the white area is (16pi r²)/4 or 4pi r² this is the same as the pink area in the square above. We can see from this that half of the left rectangle is pink. Or we can continue with our algebra. The area of each square is 16r² (its sidelength is 4r) so the pink but here is 16r² – 4pi r², so if we add this to the bit above it we have 16r² shaded (and 16r² white).

The top right has 4 shaded quarter circles, each with radius 2r, so the total shaded area is 4pi r². Below it the white circle has the same radius so same area 4pi r², again that makes the shaded area 16r² – 4pi r² so the shaded area in the right rectangle 16r², and in the big square 32r². The total area of the big square if 64r² (radius is 8r) so the shaded fraction is a half. Which is nice.

Now I know I started by saying I thought it looked like a half, which is what I did think at the time I first saw it. But I still think it’s a surprise fraction. I’ve done a lot of geometry puzzles. And many have included shapes like this, so when I look at this that knowledge helps me see. Most students would not have done anywhere near the amount of puzzles I have so wont have that foresight, and I think it would be a very surprising result that could open many up to the wow factor. I also think that it might be a good starting point for some rich class discussions.

I think it’s a great visual and a lovely answer.

Algebraic sequences puzzle

June 18, 2020 Leave a comment

I came across another Mr Gordon (@MrGordonMaths) brainteaser that I liked the look of:

It looked interesting to me. It’s not a type of sequence problem I’ve seen lots before and I thought it worth exploring.

I considered the first 2 terms. From this I can form an equation:

x² – 24x +144 = (3x -2)/7

Which leads to

7x² – 168x + 1008 = 3x -2

7x² – 171x +1010 = 0

Which facorises quite nicely

(7x-101)(x-10)

I expect that means x = 10 is our solution as surely the intention is an integer sequence.

This would give us -2, 4, 16 as our known terms. So we know that

10a+ 6 = 16²

10a + 6 = 256

10a = 250

a = 25. Which is a nice answer and a nice solution.

Although it doesn’t actually specify integer answers, so what happens if we use x = 101/7 ?

17/7 , 289/49 , 317/14 son the third term doesn’t work. So this x value doesn’t generate a valid solution.

I then wondered what would happen if I used the second and third term to generate an equation:

((3x-2)/7)² = (3x+2)/2

2(3x-2)² = 49(3x+2)

18x² – 24x + 8 = 147x +98

18x² – 171x – 90 = 0

(18x +9)(x-10) = 0.

This one again generates 2 solutions for x, but only one matches the other solution. So the common x value must be the only solution.

This was a lovely puzzle that I enjoyed thinking and working through. If you have a different solution I’d love to see it.

Circles and an octagon

June 15, 2020 Leave a comment

Here’s an interesting puzzle that came via Diego Rattaggi (@diegorattaggi) and involves circles and octagon.

I started as always with a diagram:

I labelled some sides up, then changed my mind and changed labels as I was thinking about taking a coordinate geometry approach and didn’t want to have used x and y. As it happened I didnt use that approach anyway. While looking at the sketch I realised that the triangle AOC was a right angled isoceles. Due to an error a few weeks back I wanted to double check I wasn’t making an assumption here so did some work to justify this was the case:

I was using some similar reasoning to this hexagon puzzle, I could justify that to had to be an isoceles, and that extending the lines gave an isoceles, I could just that the vetex was definitely on diameter I’d drawn and was equidistant from both circles in x and y, but felt that wasn’t enough, and if it wasn’t definitely the centre there could be multiple solutions, then I saw a different version in my screenshot:

Once I had this information it was fairly straightforward using Pythagoras’s Theorem:

At this point I realised the ratio if the radii squared was the same as the area so that’s all I needed:

I got to the end and realised I had my fraction upside down so I flipped it over.

This was an interesting puzzle, and I think I will need to think further on the case where the centre being the exact of the right triangle wasn’t specified. I might need to look on geogebra.

Circles on a line

June 12, 2020 2 comments

I saw this lovely question from Mr Gordon (@MrGordonMaths) the other day:

I looked at it and even though it said it was GCSE maths only it didn’t look at all obvious how to find an answer. It did look interesting though, I wondered how my y11s would get on with it. I thought I’d give it a try:

As always I drew a sketch:

I was looking at straight line shapes I could draw and realised the trapezium was the better option in this case:

From here it was a bit of Pythagoras’s theorem:

Which gave me all I needed for the final answer, which is 1:4. (Obviously I discounted the trival R=0 as it doesn’t make sense in the context of the question).

A nice little puzzle that I can see could be rather taxing for students despite using only concepts they will have learned in KS3 and 4. It’s the sort of question that can really help with problem solving ability and is one i will definitely try on my year 11s when we are back fully.

I’d love to see how you solved this one, especially if you took a different approach.

A great 1-9 puzzle

June 11, 2020 1 comment

This number puzzle was one I really enjoyed and it came from @1to9puzzle :

When I looked at it I did think about setting up 8 equations with 8 unknowns and solving them as one big system. But then I figures there was probably a better way.

I looked at the sums and decided that the one summing to 10 would be a good place to start. That means I need 2 numbers that sum to 6. Which gives 1 and 5 (as 4 is already taken and we can only have 1 of each number). I knew the 2 on this diagonal needed to be 1 and 5 but wasnt sure which way round they were yet.

Then I wrote some number bonds to different numbers down. I considered the middle horizontal row. It needed to sum to 12. 9 and 3, 8 and 4, 7 and 5, 6 and 6. It couldn’t be 6 and 6 as I was only allowed 1 if each. I knew the 4 and 5 were already taken on the diagonal and in the centre so this line had to be 9 and 3.

The 5 and 3 couldn’t be on the right side together as if they were on the right that would leave me needing 10 to make 18. The 5 and 9 on this side would mean I needed 4 to make 18, so I needed the 1 in the bottom right. If I then had the 3 above it I’d need to add 14 but only had 1 more square so that meant I needed to put the 9 in the middle right. From there it was a case of simple addition and subtraction to fill in the rest:

I really enjoyed this one. Would love to hear how you did it, and do send me any others like this you find.

Categories: #MTBoS, KS2, KS3, Starters, Teaching Tags: , , ,

A hexagon and some interior lines

June 8, 2020 Leave a comment

Today’s puzzle comes from Eylem Gercek Boss (@_eylem_99) and it’s a nice quick one that I loved, and includes a hexagon, which an awful.lot of puzzles I find at the moment seem to do!:

Initially there wasn’t an obvious solution to me so I sketched it put and labelled a load of things.

Then started writing what I knew:

I had 3 parallel lines equally spaced, so I had 2 similar triangles. I knew the diagonal was double the side length. I had enough to form an eqution:

2x = (1/2)x + 12

3x = 24

x = 8

From here I could easily work out the area:

A nice little problem that got me thinking about problem solving. I didn’t see a solution immediately, although perhaps I should have, so I just started jotting down what I knew until I saw a way forward. This is a key still that students need, just being able to consider what they do know an look at what that means in the context of the question. I think this question would be really good to use with students to model that thought process.

Do you have a different solution? I’d love to hear it.

Equilateral triangles in a semi-circle

June 2, 2020 1 comment

Today’s puzzle comes from Diego Rattaggi (@diegorattaggi):

It’s a nice looking puzzle so I thought I’d have a crack.

I started by drawing a truly terrible diagram:

I drew the 2 radii and managed to set up 2 equations:

I then realised I had enough information to express y in terms of x:

Which in turn meant I had 2 equations in 2 variables, so I could solve them:

And once I had r I could calculate the area nicely.

This was a nice little puzzle. I’d love to hear how you solved it.

Circles, hexagons and a lot of dead ends

May 21, 2020 Leave a comment

Today’s puzzle comes from @Catriona Shearer (@cshearer41) and is kinda great:

Have a go, before reading on.

I know sometimes when I type up my solutions it looks like the answer comes quickly and easily to me. And this is sometimes the case, but not always. And this was one of those time where it really didn’t. Here are some of the pages of dead end

And there were many more over a couple of sittings.

Today I looked with fresh eyes and considered what I had. I knew I had 2 overlapping congruent regular hexagons to make the whole shape (I will referto a hexagon this size as H1). I also knew that the white intersection was made up of a regular hexagon around a circle (H2) and 2 trapezia around semi-circles that would also make a regular hexagon (H3). (If you can’t see that these are regular consider symmetry, the angles you know are 120 and equal tangents).

I knew that the side lengths of H1 was equal to the sum of the side lengths of H2 and H3, so I figured if I could express the side length of H2 in terms of H3 or vice versa I would be able to solve it quite easily. I then had the realisation that side length of H3 had to be half of the side length of H2. Although given that I’d inferred something wrongly the other day I thought I’d better justify it fully to make sure:

Once it was justified the problem was fairly straightforward.

Side length ratio H1:H2:H3 is 3/2:1:1/2

So area ratio is: 9/4:1:1/4

The whole shape area is 2H2 – H2 -H3 so 13/4.

The white area is H2+H3 so 5/4

So the white area as a fraction of the whole is 5/13 hence the yellow area as a fraction of the whole is 8/13.

A lovely solution in the end, but one that came after a lot of dead ends. A very enjoyable puzzle, as all Catriona’s are

%d bloggers like this: